Chapter 9, Problem 82P

(a)

To determine

Find the air pressure difference between the top and bottom of the Willis tower.

Answer to Problem 82P

The air pressure difference between the top and bottom of the Willis tower in Pascal is $5.2\text{kPa}$ and in atmospheric condition is $0.051\text{atm}$.

Explanation of Solution

Write the equation for pressure difference.

$\Delta P=\rho gh$ (I)

Here, $\rho$ is the density of the air at $20°\text{C}$, $g$ is the gravitational acceleration, $h$ is the height of the Willis tower, and $\Delta P$ is the pressure difference between the top and bottom of the Willis tower.

Conclusion:

Substitute $1.20{\text{kg/m}}^3$ for $\rho$, $9.80{\text{m/s}}^2$ for $g$, and $440\text{m}$ for $h$ in equation (I).

$\begin{array}{c}\Delta P=\left(1.20{\text{kg/m}}^3\right)\left(9.80{\text{m/s}}^2\right)\left(440\text{m}\right)\\ =5174.4\text{Pa}\\ =5.2\times {10}^3\text{Pa}\left(\frac{{10}^{-3}\text{kPa}}{1\text{Pa}}\right)\\ \simeq 5.2\text{kPa}\end{array}$

Thus, the pressure difference between the top and bottom of the Willis tower in Pascal is $5.2\text{kPa}$.

Convert pressure difference in atm.

$\begin{array}{c}\Delta P=5.2\times {10}^3\text{Pa}\left(\frac{1\text{atm}}{1.013\times {10}^5\text{Pa}}\right)\\ =5.13\times {10}^{-2}\text{atm}\\ =0.051\text{atm}\end{array}$

Thus, the pressure difference between the top and bottom of the Willis tower in atmospheric condition is $0.051\text{atm}$.so it is less at the top of the tower.

(b)

To determine

Find the number of pressure decreases in Pa for every meter of altitude.

Answer to Problem 82P

The number of pressure decreases in Pa for every meter of altitude is $11.8\text{Pa/m}$.

Explanation of Solution

Write the equation for pressure difference.

$\Delta P=\rho g\Delta y$ (II)

Here, $\Delta y$ is the altitude of the tower in which the pressure decreases.

The number of Pa the pressure decreases for every meter of altitude is.

$\frac{\Delta P}{\Delta y}=\rho g$ (III)

Conclusion:

Substitute $1.20{\text{kg/m}}^3$ for $\rho$ and $9.80{\text{m/s}}^2$ for $g$ in equation (III).

$\begin{array}{c}\frac{\Delta P}{\Delta y}=\left(1.20{\text{kg/m}}^3\right)\left(9.80{\text{m/s}}^2\right)\\ =11.76\text{Pa/m}\\ \simeq \text{11}\text{.8}\text{Pa/m}\end{array}$

Therefore, the number of pressure decreases in Pa for every meter of altitude is $11.8\text{Pa/m}$.

(c)

To determine

Find the altitude at which the atmospheric pressure reaches zero.

Answer to Problem 82P

The altitude at which the atmospheric pressure reaches zero is at $8.61\text{km}$.

Explanation of Solution

Write the equation for pressure variation with depth.

$P_2=P_1+\rho gd$ (IV)

Here, $P_2$ is the pressure at the surface point 2, $d$ is the depth, and $P_1$ is the pressure at the surface point 1.

If the object is kept on the atmosphere, suppose we take point 1 at the surface and point 2 is a depth, then $P_1=P_{\text{atm}}$ so the pressure at a depth below the surface equation (IV) can be written as,

$P_2=P_{\text{atm}}+\rho gd$ (V)

Substitute 0 for $P_2$ and $-h$ for $d$ in the equation (V) and solve for $h$.

$\begin{array}{c}{P}_{\text{atm}}-\rho gh=0\\ P_{\text{atm}}=\rho gh\\ h=\frac{P_{\text{atm}}}{\rho g}\end{array}$

Conclusion:

Substitute $1.20{\text{kg/m}}^3$ for $\rho$, $1.013\times {10}^5\text{Pa}$ for $P_{\text{atm}}$, and $9.80{\text{m/s}}^2$ for $g$ in above equation.

$\begin{array}{c}h=\frac{\left(1.013\times {10}^5\text{Pa}\right)}{\left(1.20{\text{kg/m}}^3\right)\left(9.80{\text{m/s}}^2\right)}\\ =0.0861\times {10}^5\text{m}\\ =8.61\times {10}^3\text{m}\left(\frac{{10}^{-3}\text{km}}{1\text{m}}\right)\\ =8.61\text{km}\end{array}$

Therefore, the altitude at which the atmospheric pressure reaches zero is at $8.61\text{km}$.

(d)

To determine

Find the true or false statement

Answer to Problem 82P

The true statement, if the pressure is non-zero at a certain altitude and the atmosphere extends to a higher altitude.

Explanation of Solution

Since the zero altitude is calculated from an expression inverse proportionality to the air density, so that decreasing air density means that the atmospheric extends to a higher altitude.